What is the integral of e^{ct} ?

The integral of e^{ct} is 1/c e^{ct}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^{ct}

\int e^{c t} d t

\frac{1}{c} e^{c t} + C

Solution steps

\int e^{c t} d t

Apply u-substitution

= \int \frac{e ^{u}}{c} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{c} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{c} e^{u}

Substitute back

u = c t

= \frac{1}{c} e^{c t}

Add a constant to the solution

= \frac{1}{c} e^{c t} + C

\int \frac{2}{5 x} - \frac{3}{x ^{2}} d x

\int \frac{6 x ^{2}}{x ^{3} - 2} d x

\int 2^{t a n (x)} sec^{2} (x) d x

\int \frac{sin ^{2} ( 2 x )}{cos ( 2 x )} d x

\int (\frac{x ^{3} + 1}{x ^{2}}) d x